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umka21 [38]
3 years ago
13

What is a system of equations for the following situation? There is a total of 40 questions on a test. Some questions are multip

le-choice and some are short answers. The multiple-choice questions are worth 2 points and the short answer questions are worth 4 points. There is a total of 100 points on the test.
Mathematics
1 answer:
Elden [556K]3 years ago
8 0

First we define the variables:

m = multiple choice questions

s = short answer questions

Next we need to satisfy 2 conditions:

1) There are a total of 40 questions. We can represent this mathematically using the variables I defined:

m+s=40

2) The multiple-choice questions are worth 2 points and the short-answer are worth 4 points but they need to total 100 points. This can also be represented mathematically:

2m+4s=100

By satisfying the conditions provided we were able to create the 2 equations!

m+s=40

2m+4s=100

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Fittoniya [83]

Answer:

-6/16

Step-by-step explanation:

<u>Step 1:  Find equivalent to -3/8</u>

(-3*2) / (8*2)

<em>-6/16</em>

Answer:  -6/16

7 0
3 years ago
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This is Matrix for pre calc
gavmur [86]

The given system of equations in augmented matrix form is

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\-6&1&2&4&-12\\1&-3&-3&5&-20\\-2&5&6&0&12\end{array}\right]

If you need to solve this, first get the matrix in RREF:

  • Add 2(row 1) to row 2, row 1 to -3(row 3), and 2(row 1) to 3(row 4):

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&11&5&-13&37\\0&19&10&4&-10\end{array}\right]

  • Add 11(row 2) to -5(row 3), and 19(row 1) to -5(row 4):

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&-164&132&-1052\end{array}\right]

  • Add 164(row 3) to -91(row 4):

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&13080&-39240\end{array}\right]

  • Multiply row 4 by 1/13080:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&1&-3\end{array}\right]

  • Add -153(row 4) to row 3:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&0&-364\\0&0&0&1&-3\end{array}\right]

  • Multiply row 3 by -1/91:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Add 6(row 3) and -8(row 4) to row 2:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&0&0&-10\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Multiply row 2 by 1/5:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Add -2(row 2), 4(row 3), and -2(row 4) to row 1:

\left[\begin{array}{cccc|c}3&0&0&0&3\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Multiply row 1 by 1/3:

\left[\begin{array}{cccc|c}1&0&0&0&1\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

So the solution to this system is (w,x,y,z)=(1,-2,4,-3).

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Vadim26 [7]

Answer:

B

Step-by-step explanation:

The answer is B, -12.

Please give me brainliest.

-12^2 --> -144

-144+144 = 0

Therefore, the answer is B, -12.

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zmey [24]
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Art [367]

Answer:

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Step-by-step explanation:

Multiply by 2/2 to remove the decimal.

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